I'm working on the proof of the "$H=W$" theorem in the book SOBOLEV SPACES by Adams and Fournier,
and there is an argument that seems quite strange to me. Let me introduce some conventions first. $H^{m,p}(\Omega)$ is defined as the completion of
$$\{u\in C^m(\Omega):\lVert u\rVert_{m,p}<\infty\}$$
w.r.t. the norm
$$\lVert u\rVert_{m,p}=\left(\sum_{|\alpha|\leq m}\lVert D^\alpha u\rVert_p^p\right)^\frac{1}{p},$$
where $D^\alpha$ is understood in the weak sense, while $W^{m,p}(\Omega)$ is defined by weak derivatives in the usual way. As to completions, the authors mentioned in the preliminary chapter that every normed space $X$ is either a Banach space or a dense subset of a Banach space $Y$ called the completion of $X$ whose norm satisfies
$$\lVert x\rVert_Y=\lVert x\rVert_X$$
for every $x\in X$. Finally, throughout our discussion, $\Omega$ denotes a nonempty open subset of the Euclidean space. Now let us see the proof to be understood.
The argument underlined with red is much annoying. I don't see any relevance between completeness of $W^{m,p}(\Omega)$ and the extension to the asserted isometric isomorphism. If the identity operator on $S$ means the inclusion of $S$ in $W^{m,p}(\Omega)$, how could I extend it to the isometric isomorphism? Before digging into the proof, I have reviewed the preliminary chapter, so I know the normed space $S$ is isometrically isomorphic to a dense subspace of a Banach space $H^{m,p}(\Omega)$? But this doesn't seem to be any helpful. How could I use this fact to build the extension. Help is much needed. Thank you for your precious time.
Update 1: I remember that if the codomain $T$ of a uniformly continuous map $f:A\subseteq M\to T$ is complete, then we can uniquely extend $f$ to the closure $\bar{A}$, preserving uniform continuity. But this seems to give nothing if I extend the inclusion of $S$ in $W$.
Update 2: Though it may be obvious to people who know about completions of normed spaces, I'd like to emphasize for once that every normed space $X$ is in fact isometrically isomorphic to a dense subspace of a Banach space and it is this Banach space that is defined as the completion of $X$. Then, by identifying isomorphic spaces, the authors conclude that every normed space is a dense subset of its completion, I guess.
Best Answer
Let's start from $S\subseteq W^{m,p}(\Omega)$.
Consider the closure $\bar{S}$ of $S$ with respect to the topology on $W^{m,p}(\Omega)$ induced by its norm.
Claim 1: $\bar{S}$ is a linear subspace of $W^{m,p}(\Omega)$.
Proof. Check that $\bar{S}$ is closed under addition and scalar multiplication.
Claim 2: $\bar{S}$ is complete (hence it's a Banach space).
Proof. If you take a Cauchy sequence in $\bar{S}$, then it is Cauchy also in $W^{m,p}(\Omega)$, as the norm we are considering on $\bar{S}$ is the same as the one of $W^{m,p}(\Omega)$. Since $W^{m,p}(\Omega)$ is a Banach space, our Cauchy sequence needs to converge in $W^{m,p}(\Omega)$, but as $\bar{S}$ is closed, it must converge actually in $\bar{S}$.
Claim 3. $\bar{S}$ is the completion of $S$.
Proof. $S$ is clearly dense in $\bar{S}$, which is a Banach space by our second claim. Thus $\bar{S}$ is a completion of $S$, but it is also the unique one up to isometric isomorphism (because in general the completion of a normed vector spaces is unique up to isometric isomorphism). Thus we can identify $H^{m,p}(\Omega)$ with $\bar{S}.$
Hence $H^{m,p}(\Omega)\subseteq W^{m,p}(\Omega)$.
The step underlined in red is saying exactly this, with a different phrasing.